\[\sum_{n=1}^{7} 2(-2)^{n-1}\]
Answer: D) cube root of 16
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Explanation:
The rule we use is
![x^{m/n} = \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D)
In this case, x = 4, m = 2 and n = 3.
So,
![x^{m/n} = \sqrt[n]{x^m}\\\\\\4^{2/3} = \sqrt[3]{4^2}\\\\\\4^{2/3} = \sqrt[3]{16}\\\\\\](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%5C%5C%5C%5C%5C%5C4%5E%7B2%2F3%7D%20%3D%20%5Csqrt%5B3%5D%7B4%5E2%7D%5C%5C%5C%5C%5C%5C4%5E%7B2%2F3%7D%20%3D%20%5Csqrt%5B3%5D%7B16%7D%5C%5C%5C%5C%5C%5C)
Showing that the original expression turns into the cube root of 16.
Simply remember PEMDAS:
P = Parentheses
E = Exponents
M = Multiplication
D = Division
A = Addition
S = Subtraction
You perform these actions from left to right.
However in this case it is not possible for it to equal 60.
Answer:
there is no picture so it is impossible to answer
Step-by-step explanation:
There are 2 ways you can write this.
You can either do /1 or /100

The most used is